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**Neha****Member**- Registered: 2006-10-11
- Posts: 173

solve the system of equations by the substitution method:

a + 1/2b = 16

2a + b = 50

ok i know how to do equations such as these but this has a fraction......so help

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**pi man****Member**- Registered: 2006-07-06
- Posts: 251

Multiply both sides of the 1st equation by 2:

Now subtract your new first equation from your second:

Multiply both sides by b:

You can use the quadratic formula to solve this now.

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**Neha****Member**- Registered: 2006-10-11
- Posts: 173

ok thanks

i haev do the quadratic formula....

and stoped here

x = 18 +- sqrt(-320) / 2

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**Neha****Member**- Registered: 2006-10-11
- Posts: 173

hello guys what to do next......

18 +- sqrt(-320)

x = -------------------

2

help

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**Neha****Member**- Registered: 2006-10-11
- Posts: 173

I THINK YOU WRONG

a+1/2b=16

2a+b=50

2a=50-b

a=(50-b)/2

substituting in eqn. 1

(50-b)/2+1/2b=16

multiplying by 2b

b(50-b)+1=32b

50b-b^2+1-32b=0

-b^2+18b+1=0

=>b^2-18b-1=0

b=[18+/-rt(324+4)]/2

=[9+/-rt82]

similarly express b interms of a and find a

or substitute the value of a

i have a feeling your sum might be

a+(1/2)*b=16

and 2a+b=50

these are the equations of parallel lines

and hence no solution

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

b!=0

*Last edited by krassi_holmz (2006-11-11 13:47:43)*

IPBLE: Increasing Performance By Lowering Expectations.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

a + 1/2b = 16

2a + b = 50

Twice it was assumed that it is 1/(2b) in the first equation, where as I'm pretty sure it is in fact (1/2)b.

Normally students are asked to solve systems of linear equations.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

If it's (1/2)b:

No solution.

IPBLE: Increasing Performance By Lowering Expectations.

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